You are designing a pulse transformer, so the generator is appling a DC voltage to the transformer for a fixed time of half the frequency period and then reversing the polarity for the next half period and so on.

During the time when current is applied to the inductor, the inductive current will go from 0 to an amount limited only by the source impedance and the winding resistance. The inductance of the transformer primary should be large enough to ensure the inductive reactance does not become significant with respect to the source impedance of 50Ω. A current equivilent to 500Ω is a good start. This amount will cause the pulse top to droop only 10% or so.

So for (2) just ensure a large enough inductance is used to prevent the winding resistance from shorting the generator signal.

So for (2) just ensure a large enough inductance is used to prevent the winding resistance from shorting the generator signal.

Click to expand...

Do you mean when testing it under o.c. or s.c. conditions?

I don't have the inductance of the coil. Rather I am modelling it using the non-ideal model for a transformer (see the attachment). R1 and R2 are the series resistances of the primary and secondary coils, and L1 and L2 (X2) are the leakage inductances (which I think I can ignore for now?). The other branch is the magnetising branch.

Now the only things I can find at this point in time is the total series wire resistance (depending of course on the number of turns...). But of course, this value is absolutely no where near even 50 ohms, even if I have like 100 turns!

Thx for the reply.. The only prob I have is that you're modelling the transformer as two inductors with mutual inductance (a method which I'm not too familiar with), whereas I'm modelling it using the non-ideal equivalent circuit for a transformer. But anyway....

If you're only considering the 50Ω resistor, then P = V^2/R, so V = √50 = 7.07V. But then isn't there also power lost in the transformer itself (windings have resistance)?

I = V/R = 0.141A

X = 2∏fL1
500 = 2∏(50)L1
L1 = 1.59H

I'm guessing L2 = L1*3 = 4.77H. I'm not sure?

Open circuit and short circuit.

I dunno Never simulated something like this before, and don't plan to... I was just using Multisim it to draw the circuit lol. Paper and test method will have to suffice

Ok so now that I have (I think) the inductances, I'm guessing that L = N^2 / R, where R is reluctance. So then I need the reluctance of the core? And what was the point of finding current in the first place?

Designing transformers is a bit of an art. More than one solution is usually possible. Being too critical in your design can be costly. Arbitray decisions about what is required is common because there can be more than one good answer. It's not like logic design where everything has to be a one or a zero, results half to be correct and nothing in-between is allowed. It takes years of experience to get good at it.

It is desired that transformer, like this one, transform impedances with out losses, or distorting the signal. The reality is that there will be losses and distortion, but they must be at a level you can live with.

Ignore the losses for now.

Good. The RMS current through the un-loaded transformer primary should one tenth of this value. Δ XL = 50Ω x 10 = 500Ω. This is a somewhat arbitrary value. It certainly won't overload the gen.

Ok thx, but just this unloaded story is confusing me. By unloaded, you no 50Ω or no R_L? When I said o.c. or s.c. I meant, testing with the output o.c. or s.c. ..

Oh sorry So used to working with 50Hz! L = 1mH now...much better

L = N^2μA/l . I've got L, A and l, so I can then get N. I get N = 10.1 ≈ 11 turns minimum. Anything obove this will be even better right?

I won't be simulating, so once I've got N_prim, then N_sec = N_prim*3.. or use inductance ratio (thx for the link). N_sec = 22 turns

Yes - I've got the core and everything. I used values from the datasheet to calculate the above.

For the rest of the calculations, I'm still confused with this:

Why do you use 1W for the power consumption of the 50Ω internal resistor AND the load resistor. Doesn't the VA rating give you the rated current, namely, for the primary: I_rated = 1 / 10 = 0.1Arms (since RMS voltage = peak value for square waves)

I mean no 50Ω. However, when you finally do terminate the secondary of the transformer, it will be a value higher than 50Ω. Calculate that value.

Good. 1mH is a good starting value and if the pulse looks good to you then stay with that value.

From what you say, you get approximately 1mH with 11 turns on the core you have. More inductance is better, but inter-winding capacitance and winding resistance goes up as well. Try 11 turns for now.

N_sec = N_prim*3. So how many turns are needed on the secondary? What value of resistor across the secondary do you need to properly terminate the generator? Read and calculate.

What size of wire are you going to use?

You allready calculated the generator output will be set to 7.07V RMS. That's 0.141 Amps RMS through a 50Ω impedance the generator sees. That impedance is reflected from the secondary terminating impedance... which you haven't calculated yet.

The signal is a square wave so the peak current is 0.141 Amps. The core material that you choose should not saturate with this value of current. This is all based on the original requirement of 1 VA for the transformers power handling capability. Do the core specs say anything about power or VA rating?